I Using position of free particle to measure time

Kashmir
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Hartle, Gravity

"An observer in an inertial frame can discover a parameter ##t##with
respect to which the positions of all free particles are changing at constant rates.
This is time"

Then goes on to say
"Indeed, inertial frames
could be defined as Cartesian reference frames for which Newton’s first law holds
in the form ##\mathbf{\ddot r}=0##
Using the laws of mechanics, an observer in an inertial frame can construct a
clock that measures the time ##t##. For instance, the position of one free particle could
be used to measure ##t##, since its position changes at a constant rate in ##t##"

Using the position of one free particle to measure time, how does that guarantee that for other free particles this definition of time leads to ##\mathbf{\ddot r}=0##
 
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Kashmir said:
Using the position of one free particle to measure time, how does that guarantee that for other free particles this definition of time leads to ##\mathbf{\ddot r}=0##
It doesn’t. We still have to observe the other particles to verify that ##\mathbf{\ddot r}=0## holds for them as well; if we find exceptions then we know that we don’t have an inertial frame.
 
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You need three particles moving in uniform motion in non-coplanar directions to establish an inertial frame of reference.
 
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Kashmir said:
Using the position of one free particle to measure time, how does that guarantee that for other free particles this definition of time leads to ##\mathbf{\ddot r}=0##
It does not. An explicit counter example is a rotating reference frame where the one free particle is moving along the axis of rotation. As @vanhees71 you need multiple particles to accomplish this.
 
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@vanhees71 then what does the author mean by constructing a clock using a free particle ? Can you please explain it? Thank you
 
Nugatory said:
It doesn’t. We still have to observe the other particles to verify that ##\mathbf{\ddot r}=0## holds for them as well; if we find exceptions then we know that we don’t have an inertial frame.
What does the author mean then by constructing a clock using a free particle? Can you please explain it. Thank you :)
 
Kashmir said:
Using the position of one free particle to measure time, how does that guarantee that for other free particles this definition of time leads to ##\mathbf{\ddot r}=0##
You are looking at it backwards. You have to establish that you have an inertial frame (which requires more than one free particle) first. Then, once you have done that, you can pick anyone of the free particles and use its position to define time.
 
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Kashmir said:
What does the author mean then by constructing a clock using a free particle? Can you please explain it. Thank you :)
Anything that changes can be used as a clock: in the past we’ve measured time with sand moving through a sandglass, the length of a burning candle, the shadow of the sun, nowadays we sometimes use the rate of decay of radioactive materials…. Why not use the position of a moving object as a clock? Come to think of it…. Is that not what’re doing with an ordinary dial clock or stopwatch? The hands are moving, we tell the time by where they’ve moved to.
 
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PeterDonis said:
You are looking at it backwards. You have to establish that you have an inertial frame (which requires more than one free particle) first. Then, once you have done that, you can pick anyone of the free particles and use its position to define time.
To establish you've an inertial frame you need that for free particles there is no acceleration i.e ##\ddot r=0##. so you already have defined time to prove the frame is inertial. Isn't your logic circular?
 
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What’s wrong with circular logic? It is certainly self consistent.
 
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Nugatory said:
Anything that changes can be used as a clock: in the past we’ve measured time with sand moving through a sandglass, the length of a burning candle, the shadow of the sun, nowadays we sometimes use the rate of decay of radioactive materials…. Why not use the position of a moving object as a clock? Come to think of it…. Is that not what’re doing with an ordinary dial clock or stopwatch? The hands are moving, we tell the time by where they’ve moved to.
Author says "An observer in an inertial frame can discover a parameter ##t## with respect to which the positions of all free particles are changing at constant rates.
This is time"To establish I've a inertial frame I've to find that free particles move in straight lines and the double derivative with respect to a parameter, call it ##a##, is zero for all the particles.
So I've to search for a parameter ##a## that gives ##d^2 r/da =0## for all the free particles.

The author chooses ##a## as related to the distance of one particular free particle and says this can be used to measure time.
But how does one know that we choose the correct paramete?

We were guaranteed that there exists a parameter such that ##d^2 r/da =0## but it didn't tell me that the position of one free particle is a correct parameter.
 
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Kashmir said:
To establish you've an inertial frame you need that for free particles there is no acceleration i.e ##\ddot r=0##. so you already have defined time to prove the frame is inertial. Isn't your logic circular?
This physical determination of an inertial reference frame goes back to Lange (1885). The idea is that you can establish an inertial frame by taking one free particle, moving in a straight line relative to the frame to define the time measure via measurements of the distance traveled by this particle. In addition you then need to establish that with this measure of time three free particles moving in non-coplanar directions in fact move with constant velocity to that reference frame.

For a translation of this paper, see

https://link.springer.com/content/pdf/10.1140/epjh/e2013-40040-5.pdf
 
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Kashmir said:
But how does one know that we choose the correct parameter?
what would be an “incorrect” parameter? And is it possible for the position of a free particle to be one?
 
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vanhees71 said:
This physical determination of an inertial reference frame goes back to Lange (1885). The idea is that you can establish an inertial frame by taking one free particle, moving in a straight line relative to the frame to define the time measure via measurements of the distance traveled by this particle. In addition you then need to establish that with this measure of time three free particles moving in non-coplanar directions in fact move with constant velocity to that reference frame.

For a translation of this paper, see

https://link.springer.com/content/pdf/10.1140/epjh/e2013-40040-5.pdf
Thank you. This is what I was asking for.
I don't have access to the article but I think your comment did resolve much of my doubt .

But is it sufficient that the measure of time you define, related to distance of a free particle, which leads to zero accelerations for three particles moving along non coplanar paths is sufficient enough that all other free particles will have zero accelerations as well if we use this time?
 
  • #15
Nugatory said:
what would be an “incorrect” parameter? And is it possible for the position of a free particle to be one?
An incorrect one would be that the accelerations are non zero for free particles.

I'm not sure about the second part.
 
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Dale said:
What’s wrong with circular logic? It is certainly self consistent.
Please see post #11 above. I've written more clearly about my doubt. Thank you :)
 
  • #17
Kashmir said:
But how does one know that we choose the correct paramete?
Usually you just plug the parameter back into the definition and see if it works. If you find that for some free particle ##d^2 r/da^2 \ne 0## then you know the choice is incorrect. Otherwise it is correct.

This is very circular, but it is also correct.
 
  • #18
Kashmir said:
An incorrect one would be that the accelerations are non zero for free particles.
Then we know we have made a poor choose of parameter, so we try choosing another one that works better.

Start with Einstein's "Time is what a clock measures"; anything that parameterizes the passage of time can be used as a clock. We generally try to choose our clocks so that ##\mathbf{\ddot r}=0## holds for free particles.
 
  • #19
Kashmir said:
But is it sufficient that the measure of time you define, related to distance of a free particle, which leads to zero accelerations for three particles moving along non coplanar paths is sufficient enough that all other free particles will have zero accelerations as well if we use this time?
@vanhees71 . Could you comment on it?
 
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